This problem is present in "Supplements to the Exercises in Chapter 1-7 of Walter Rudin's Principles of Mathematical analysis" by Prof. George M. Bergman, which states as follows,
Show that if $(a_{n})$ is a non-absolutely convergent series, and $\alpha$ is a real number, then there exists a rearrangement $(a_{k_{n}})$ of $(a_{n})$ such that for all even $n$, $a_{k_{n}} \geq 0$, for all odd $n$, $a_{k_{n}} \leq 0$, and $\sum a_{k_{n}} = \alpha$.
This problem looks like the Riemann series theorem with constraint in the series. Can anyone give me some clues about how to construct it?
What I have got is that: first, this series must contain infinitely many positive and negative terms; second, the problem can be boiled down to this one: if the conditional series $(a_{n})$ converges to zero, find a rearrangement which also converges to zero and such that its odd terms are $\leq 0$, and its even terms are $\geq 0$. However, I stuck here for a long time.
Thanks in advance.
Such a rearrangement can be constructed in a stage-wise manner as we show below. Let $(b_n)$ denote the subsequence of all non-negative numbers from $(a_n)$ and let $(c_n)$ denote the subsequence of all negative numbers from $(a_n)$. As you had indicated in the question, both $(b_n)$ and $(c_n)$ are non-empty infinite sequences since $(a_n)$ is conditionally convergent. Let us also observe the following facts:
Now, we construct the required rearrangement in a stage-wise manner by inspecting the sequences $(b_n)$ and $(c_n)$. Intuitively we do the following. We inspect the sum of the finite sequence constructed until stage $k-1$, say $s_{k-1}$ and check if $s_{k-1} \geq \alpha$ or $s_{k-1} < \alpha$. If $s_{k-1} \geq \alpha$, we pair up numbers from $(c_n)$ with an insignificant subsequence of $(b_n)$ and walk backwards along the number line until the sum reaches below $\alpha$. That such insignificant subsequences exists and that the sum shall reach below $\alpha$ is guaranteed due to the observations above. Similarly, if $s_{k-1} < \alpha$, we shall pair up up numbers from $(b_n)$ with an insignificant subsequence of $(c_n)$ and walk forwards along the number line until the sum reaches above $\alpha$. The concatenation of such stage-wise walks can be shown to be a rearrangement with the required properties.
Let us start with the empty sequence at stage $0$ and let us say that all elements in $(b_n)$ and $(c_n)$ are not chosen at the start (at each stage we shall only have finitely many elements that are chosen in $b_n$ and $c_n$). At stage $k$, we us do the following:
Let $s_{k-1}$ be the sum of the finite sequence constructed in stage $k$ (where we assume that $s_0=0$). We have the following cases: either $s_{k-1} \geq \alpha$ or $s_{k-1} < \alpha$ . If $s_{k-1} \geq \alpha$, we do the following:
Let $(b_{k_n})$ be any subsequence of elements in $(b_n)$ that are not chosen until stage $k-1$ such that for any $n$, $\lvert b_{k_n} \rvert \leq \frac{1}{2^{k+n}}$. This ensures that $0 \leq \sum_n b_{k_n} \leq \frac{1}{2^k}$ (since $b_n$'s are non-negative). Such a sequence exists since $b_n \to 0$.
Let $(c_{k_n})$ be the sequence of all elements in $(c_n)$ that are not chosen until stage $k-1$.
Consider the alternating sequence $c_{k_1},b_{k_1},c_{k_2},b_{k_2},c_{k_3},b_{k_3},c_{k_4},b_{k_4}\dots$. Let $l$ be the smallest number such that such that $s_{k-1}+\sum_{i=1}^{l} (c_{k_i}+b_{k_i}) < \alpha$. Such an $l$ exists since $\sum_n c_n \to -\infty$ and $\sum_n b_{k_n} \leq \frac{1}{2^k}$.
Let us append the sequence $c_{k_1},b_{k_1},c_{k_2},b_{k_2},c_{k_3},b_{k_3} \dots c_{k_{l}},b_{k_{l}}$ to the finite sequence we constructed until stage $k-1$ and mark these numbers as chosen in $(b_n)$ and $(c_n)$.
If $s_{k-1} < \alpha$, we perform similar steps by exchanging the roles of $b$ and $c$ and ensure that the finite sequence constructed until stage $k$ has the desirable alternating negative-positive property and $s_k \geq \alpha$.
It is not hard to see that the sequence $(a_{k_n})$ constructed in this manner has the required properties. The sequence we construct is a rearrangement of $(a_n)$ since each number in $(b_n)$ and $(c_n)$ is eventually chosen at some stage and is never chosen thereafter. The alternating negative-positive property is ensured by the construction of the sequence. We have $\sum_n a_{k_n} =\alpha$ since the sums $s_k$ at the end of stage $k$ shall not away from $\alpha$ by more than $c_{k_{l}}+\sum_n b_{k_n}$ (or $b_{k_{l}}+\sum_n c_{k_n} $) which goes to $0$ as $k \to \infty$. This is because $b_n \to 0$, $c_n \to 0$ and we have also ensured that $\sum_n b_{k_n} \leq \frac{1}{2^k}$ (or $\sum_n c_{k_n} \geq -\frac{1}{2^k}$) at the end of stage $k$.